osman@sprite.DEC (Eric Osman, dtn 283-7484) (01/28/85)
What is the integral ? Or what is it from x = 0 to 1 ?
ech@spuxll.UUCP (Ned Horvath) (01/29/85)
The rationals have measure 0 -- so the integral is 0 regardless of the integration limits. =Ned=
ech@spuxll.UUCP (Ned Horvath) (01/29/85)
Sorry about that: you have to leave 0 out of the integration range, since f(0) is undefined. Presuming 0 is not in the range, the integral is indeed 0. =Ned=
ags@pucc-i (Dave Seaman) (01/29/85)
> What is the integral ? Or what is it from x = 0 to 1 ?
The answer to both questions is zero. The function is zero "almost
everywhere," which means everywhere except on a set of measure zero.
The rational numbers are a countable set, and all countable sets have
measure zero. Whenever two functions are equal almost everywhere, it
follows that they have the same integral.
The integral I am referring to is the Lebesgue integral, which is not the
one taught in beginning calculus courses. The older Riemann integral does
not exist for the function f. Whenever a function is Riemann-integrable,
it is also Lebesgue-integrable, and the integrals agree.
--
Dave Seaman ..!pur-ee!pucc-i:agswildbill@ucbvax.ARPA (William J. Laubenheimer) (01/30/85)
Yes, it is integrable. Any function which is constant almost everywhere
(i.e., except on a set of measure 0) is integrable. The integral over
any measurable set, such as the ones you gave, is 0. This result holds
for any function which is 0 a.e.
Bill Laubenheimer
----------------------------------------UC-Berkeley Computer Science
...Killjoy went that-a-way---> ucbvax!wildbillags@pucc-i (Dave Seaman) (01/30/85)
> Sorry about that: you have to leave 0 out of the integration range, since > f(0) is undefined. Presuming 0 is not in the range, the integral is > indeed 0. In the first place, f(0) is perfectly well-defined. Since 0 = 0/1, and since gcd(0,1) = 1 (which is what it means to say that a fraction is in lowest terms), it follows that f(0) = 1. (You do have to be somewhat more precise about the definition and require q>0, but that applies to all p/q, not only at zero). Second, whether f is defined at zero has nothing to do with whether the integral exists. The function could be undefined even at uncountably many points (the Cantor set, for example) and still have an integral. It is only necessary that f be defined almost everywhere. -- Dave Seaman ..!pur-ee!pucc-i:ags
matt@oddjob.UChicago.UUCP (Matt Crawford) (01/30/85)
All you people who say "yes, it is integrable ..." may be deceiving some readers. The *Riemann* integral exists only if the limit of the finite sums exists as the `mesh size' (largest difference between any two mesh points) goes to zero. I can always choose a mesh of all rational or all irrational points and get different answers. Therefore, the function is not *Riemann* integrable. And in the words of Richard Feynman: "Lebesgue, Schlemesgue!" _____________________________________________________ Matt University crawford@anl-mcs.arpa Crawford of Chicago ihnp4!oddjob!matt
ndiamond@watdaisy.UUCP (Norman Diamond) (01/30/85)
> What is the integral ? Or what is it from x = 0 to 1 ?
The function is 0 at all but countably many points. Therefore its
integral is 0.
-- Norman Diamond
UUCP: {decvax|utzoo|ihnp4|allegra|clyde}!watmath!watdaisy!ndiamond
CSNET: ndiamond%watdaisy@waterloo.csnet
ARPA: ndiamond%watdaisy%waterloo.csnet@csnet-relay.arpa
"Opinions are those of the keyboard, and do not reflect on me or higher-ups."wjafyfe@watmath.UUCP (Andy Fyfe) (01/31/85)
In article <589@oddjob.UChicago.UUCP> matt@oddjob.UUCP (Matt Crawford) writes: >All you people who say "yes, it is integrable ..." may be deceiving >some readers. The *Riemann* integral exists only if the limit of >the finite sums exists as the `mesh size' (largest difference between >any two mesh points) goes to zero. I can always choose a mesh of >all rational or all irrational points and get different answers. >Therefore, the function is not *Riemann* integrable. And in the >words of Richard Feynman: "Lebesgue, Schlemesgue!" >_____________________________________________________ >Matt University crawford@anl-mcs.arpa >Crawford of Chicago ihnp4!oddjob!matt { 1/q, x=p/q f(x) = { , f(0) = 1 (though it doesn't matter) { 0, otherwise Now f(x) is in fact Riemann integrable over [0,1]. The reason f(x) if Riemann integrable over [0,1] is the following: Choose some n>0. Then only a finite number of x make f(x) > 1/n. I can make a mesh around those points arbitrarily narrow, so that their upper sum is made less than epsilon. And the total of the remainder of the upper sums is less than (1/n)*(1-0), so the upper sum is less than (epsilon + 1/n). The lower sum is always zero, and thus, in the limit (as n goes to infinity), we get the Riemann integral to be 0. Now if f(x) = 1 on the rationals, it wouldn't be Reimann integrable. This is because any interval making up the mesh necessarily contains both an irrational and a rational, so the upper sum is always 1, while the lower sum is always 0. What do you mean by a mesh of all rational, or irrational points? --andy fyfe ...!{decvax, allegra, ihnp4, et. al}!watmath!wjafyfe wjafyfe@waterloo.csnet
nemo@rochester.UUCP (Wolfe) (02/01/85)
> The rationals have measure 0 -- so the integral is 0 regardless of the > integration limits. > > =Ned= It is Lebesgue integrable, as opposed to Riemann-Stiljes integrable. Nemo
wjafyfe@watmath.UUCP (Andy Fyfe) (02/01/85)
It is Lebesgue's Theorem that completely characterizes all bounded Riemann integrable functions (over [a,b]). It states that a bounded function f on [a,b] is Reimann integrable if and only if the set of points at which f is discontinuous has measure zero. Just in case there was still any doubt about f(x)......
matt@oddjob.UChicago.UUCP (Matt Crawford) (02/05/85)
In article <watmath.11212> wjafyfe@watmath.UUCP (Andy Fyfe) quotes me: >In article <589@oddjob.UChicago.UUCP> matt@oddjob.UUCP (Matt Crawford) writes: >>All you people who say "yes, it is integrable ..." may be deceiving >>some readers. The *Riemann* integral exists only if the limit of >>the finite sums exists as the `mesh size' (largest difference between >>any two mesh points) goes to zero. I can always choose a mesh of >>all rational or all irrational points and get different answers. >>Therefore, the function is not *Riemann* integrable. And in the >>words of Richard Feynman: "Lebesgue, Schlemesgue!" and then goes on to give a correct explanation of why the function *is* Riemann integrable. I stand by the first portion of my state- ment above (because all previous respondents had just said that it was integrable because it was a constant (zero) almost everywhere). As for the second part, well I guess I fucked up there. Sure enough, if you estimate the integral by Simpson's rule with 2^n points the estimate is (n+2)/2^(n+1) --> 0 as n-->infinity. > >Now if f(x) = 1 on the rationals, it wouldn't be Reimann integrable. >This is because any interval making up the mesh necessarily contains >both an irrational and a rational, so the upper sum is always 1, while >the lower sum is always 0. > >What do you mean by a mesh of all rational, or irrational points? > We aren't using the same definition, but they are equivalent. I was applying the definition from my 1st year calculus course which in essence said that the limit of the sum of f(x_i)*(x_{i+1}-x_i) must go to the same constant as the max of (x_{i+1}-x_i) goes to zero, no matter how you choose the x_i's. ("_" denotes subscript.) _____________________________________________________ Matt University crawford@anl-mcs.arpa Crawford of Chicago ihnp4!oddjob!matt
jfh@browngr.UUCP (John "Spike" Hughes) (02/08/85)
In this case, the function *is* integrable, but if one sets f(x) = (if x=p/q then x else 0) you get a function that is Lebesgue integrable but not Riemann integrable. It may well be true that there is a more restictive notion of integral than Riemann integration under which functions like f(x) = (if x = p/q then 1/q\else 0) are *not* integrable. -jfh